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A parallel beam of light of wavelength $6000 Å$ gets diffracted by a single slit of width $0.3 \mathrm{~mm}$. The angular position of the first minima of diffracted light is
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The correct answer is:
$2 \times 10^{-3} \mathrm{rad}$
Given, $\lambda=6000 Å=6000 \times 10^{-10} \mathrm{~m}$ and $n=0.3 \mathrm{~mm}=0.3 \times 10^{-3} \mathrm{~m}$ $0.3 \times 10^{-3} \times \sin \theta=6000 \times 10^{-10}$
$$
\sin \theta=2 \times 10^{-3}
$$
or $\theta=2 \times 10^{-3} \mathrm{rad}$
$$
\sin \theta=2 \times 10^{-3}
$$
or $\theta=2 \times 10^{-3} \mathrm{rad}$
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